Step 1: Using a reduction identity:
\(\sin^3 2x\cos2x=\dfrac14\sin^2(2x)\cdot\sin(4x)\)... instead, directly integrate by recognizing the power rule for \(\sin2x\).
Step 2: Direct antiderivative:
\(\displaystyle\int\sin^{3}2x\cos2x\,dx=\frac{\sin^{4}2x}{8}+C\), obtained since \(\frac{d}{dx}\sin^4(2x)=4\sin^3(2x)\cos(2x)\cdot2=8\sin^3(2x)\cos2x\).
Step 3: Evaluating between the limits:
\(\Big[\frac{\sin^4 2x}{8}\Big]_0^{\pi/4}=\frac{1}{8}-0\).
Final Answer:
\[ \boxed{\dfrac18} \]